Problem: What is the value of the following logarithm? $\log_{16} \left(\dfrac{1}{256}\right)$
If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $16^{y} = \dfrac{1}{256}$ In this case, $16^{-2} = \dfrac{1}{256}$, so $\log_{16} \left(\dfrac{1}{256}\right) = -2$.